Question

(8 points) The National Assessment of Educational Progress (NAEP) tested a sample of students who had used a computer in their mathematics classes, and another sample of students who had not used a computer. The mean score for students using a computer was 309, with a standard deviation of 29. For students not using a computer the mean was 303, with a standard deviation of 32. Assume there were 60 students in the computer sample, and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ?

Answer 1

Explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let c b the subscript for students using computer and n be the subscript for students not using computer.

Therefore, the population means would be μc and μn.

The random variable is xc - xn = difference in the sample mean scores of students who used computers and those who didn't use computers

We would set up the hypothesis.

The null hypothesis is

H0 : μc = μn H0 : μc - μn = 0

The alternative hypothesis is

Ha : μc ≠ μn Ha : μc - μn ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xc - xn)/√(sc²/nc + sn²/nn)

From the information given,

xc = 309

xn = 303

sc = 29

sn = 32

nc = 60

nn = 40

t = (309 - 303)/√(29²/60 + 32²/40)

t = 0.953

The formula for determining the degree of freedom is

df = [sc²/nc + sn²/nn]²/(1/nc - 1)(sc²/nc)² + (1/nn - 1)(sn²/nn)²

df = [29²/60 + 32²/40]²/(1/60 - 1)(29²/60)² + (1/40 - 1)(32²/40)² = 1569.48/20.13

df = 78

We would determine the probability value from the t test calculator. It becomes

p value = 0.344

Assuming a level of significance of 0.05, we would not reject the null hypothesis because the p value, 0.344 is > 0.05

Therefore, we cannot conclude that the population mean scores differ.